The magnetic field of a plane electromagnetic wave is given by vec B =2 × 10-8 sin (0.5 × 103 x +1.5 × 1011 t ) hat j T The amplitude of the electric field would be

Question

The magnetic field of a plane electromagnetic wave is given by vec B =2 × 10-8 sin (0.5 × 103 x +1.5 × 1011 t ) hat j T The amplitude of the electric field would be (A) 6 Vm -1 along x-axis (B) 3 Vm -1 along z-axis (C) 6 Vm -1 along z-axis (D) 2 × 10-8 Vm -1 along z-axis

Tardigrade Admin · Accepted Answer

c =( E 0/ B 0) ⇒ E 0= cB 0 E 0=(3 × 108)(2 × 10-8) E 0=6 Vm -1 As, vec B = along y-axis vec v = along negative x-axis hence vec E 0= along z-axis

Q. The magnetic field of a plane electromagnetic wave is given by
$B = 2 \times 1 0^{- 8} sin (0.5 \times 1 0^{3} x + 1.5 \times 1 0^{11} t) \hat{j} T$
The amplitude of the electric field would be

$6 V m^{- 1}$ along $x$ -axis

$3 V m^{- 1}$ along z-axis

$6 V m^{- 1}$ along z-axis

$2 \times 1 0^{- 8} V m^{- 1}$ along z-axis

$c = \frac{E _{0}}{B _{0}} \Rightarrow E_{0} = c B_{0}$
$E_{0} = (3 \times 1 0^{8}) (2 \times 1 0^{- 8})$
$E_{0} = 6 V m^{- 1}$
As, $B =$ along y-axis $v =$ along negative $x$ -axis hence $E_{0} =$ along z-axis

Q. The magnetic field of a plane electromagnetic wave is given by B=2×10−8sin(0.5×103x+1.5×1011t)j^​T The amplitude of the electric field would be

6Vm−1 along x-axis

3Vm−1 along z-axis

6Vm−1 along z-axis

2×10−8Vm−1 along z-axis